I want to show that, when $\mu=\mu_0$, then gamma family $\Gamma(a,b)$ is a conjugate prior to inverse Gaussian with density $f(x,\mu,\lambda)=\sqrt{\frac{\lambda}{2\pi x^2}}exp[-\frac{\lambda(x-\mu)^2}{2\mu^2x}]$.
I was able to verify that $IG$ is an exponential family generated by $\bf{T}$$(X)=-1/2(X,X^{-1})^T$ and $h(x)=\sqrt{\frac{1}{2\pi x^3}}$ but I am pretty confused on how to show the conjugate prior.
Any helps would be deeply appreciated!!
$$\pi_t(\theta)=exp(\Sigma_{j=1}^k\eta_j(\boldsymbol{\theta})t_j-t_{k+1}B(\boldsymbol{\theta})-logw(\mathbf{t}))$$ $$w(t)=\int...\int\int_Rexp(\Sigma t_j\eta_j-t_{k+1}B)d\theta_1...d\theta_k$$ $$B(\theta)=-\frac{\lambda}{\mu}-\frac{log\lambda}{2}$$ Since $\mu$ is known then k=1 in this part. taking $a=t1$ and $b=\frac{t2-t1}{\mu}$, we yield: $$w(t)=\int_0^\infty exp(-b\lambda)\lambda^{a/2}d\lambda=b^{-\frac{a+1}{2}}\Gamma(\frac{a+1}{2})$$ $$\Pi_t(\theta)=\frac{\lambda^{\frac{a+1}{2}}e^{-b\lambda}}{\Gamma(\frac{a+1}{2})}b^{\frac{a+1}{2}}$$ We can conclude that this shows the gamma distribution with parameters $\alpha=\frac{a+1}{2}$ and $\beta=b$.
